Find the Domain (12y-15)/(25y^2-4)

$\frac{12 y - 15}{25 y^{2} - 4}$

Step 1

Set the denominator in $\frac{12 y - 15}{25 y^{2} - 4}$ equal to $0$ to find where the expression is undefined.

$25 y^{2} - 4 = 0$

Step 2

Solve for $y$ .

Add $4$ to both sides of the equation.

$25 y^{2} = 4$

Divide each term in $25 y^{2} = 4$ by $25$ and simplify.

Divide each term in $25 y^{2} = 4$ by $25$ .

$\frac{25 y^{2}}{25} = \frac{4}{25}$

Simplify the left side.

Cancel the common factor of $25$ .

Cancel the common factor.

$\frac{25 y^{2}}{25} = \frac{4}{25}$

Divide $y^{2}$ by $1$ .

$y^{2} = \frac{4}{25}$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{\frac{4}{25}}$

Simplify $\pm \sqrt{\frac{4}{25}}$ .

Rewrite $\sqrt{\frac{4}{25}}$ as $\frac{\sqrt{4}}{\sqrt{25}}$ .

$y = \pm \frac{\sqrt{4}}{\sqrt{25}}$

Simplify the numerator.

Rewrite $4$ as $2^{2}$ .

$y = \pm \frac{\sqrt{2^{2}}}{\sqrt{25}}$

Pull terms out from under the radical, assuming positive real numbers.

$y = \pm \frac{2}{\sqrt{25}}$

Simplify the denominator.

Rewrite $25$ as $5^{2}$ .

$y = \pm \frac{2}{\sqrt{5^{2}}}$

Pull terms out from under the radical, assuming positive real numbers.

$y = \pm \frac{2}{5}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$y = \frac{2}{5}$

Next, use the negative value of the $\pm$ to find the second solution.

$y = - \frac{2}{5}$

The complete solution is the result of both the positive and negative portions of the solution.

$y = \frac{2}{5}, - \frac{2}{5}$

Step 3

The domain is all values of $y$ that make the expression defined.

Interval Notation:

$(- \infty, - \frac{2}{5}) \cup (- \frac{2}{5}, \frac{2}{5}) \cup (\frac{2}{5}, \infty)$

Set-Builder Notation:

${y | y \neq \frac{2}{5}, - \frac{2}{5}}$

Step 4

Find the Domain (12y-15)/(25y^2-4)

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